For \(x\ge 0\)  and \(y> 0\) , it is not difficult to show that \(\frac{x^2}{y}\)  is jointly convex with respect to \(x\)  and \(y\) . Furthermore, for \(\frac{x^2}{y}\ge 0\)  and \(p\ge 1\) , it follows that \((\frac{x^2}{y})^p\)  is jointly convex with respect to \(x\)  and \(y\)  by invoking the composition rule of convexity [1]} [Chapter 3.2, Page 84]. Setting \(p=2\) , \(x= | {\bar{\mathbf {q}}}^H_k \bar{\mathbf {v}}_0 |\) , and \(y= \frac{1}{\sqrt{\tau _0}}\) , we obtain the convexity of \(\tau _0 | {\bar{\mathbf {q}}}^H_k \bar{\mathbf {v}}_0 | ^4\)  with respect to \( \bar{\mathbf {v}}_0\)  and \(\frac{1}{\sqrt{\tau _0}}\) .

Recall that any convex function is globally lower-bounded by its first-order Taylor expansion at any point. This thus motivates us to apply the SCA technique for solving problem (REF ). Therefore, with given local point \( \mathbf {w}_0\)  and \(t_0\) , we obtain the following lower bound for \(\tau _0 | {\bar{\mathbf {q}}}^H_k \bar{\mathbf {v}}_0 | ^4 = \left( { | {\bar{\mathbf {q}}}^H_k \bar{\mathbf {v}}_0 | ^2}/{ \frac{ 1}{ \sqrt{\tau _0} } } \right)^2\)  as
\(\left( \frac{ | {\bar{\mathbf {q}}}^H_k \bar{\mathbf {v}}_0 | ^2}{ \frac{ 1}{ \sqrt{\tau _0} } } \right)^2 &\ge t_0 | {\bar{\mathbf {q}}}^H_k \mathbf {w}_0 | ^4 +2 \left( \frac{ | {\bar{\mathbf {q}}}^H_k \mathbf {w}_0 | ^2}{\frac{ 1}{ \sqrt{t_0} } } \right)\left( \frac{ 2 \mathrm {Re}\lbrace  \mathbf {w}^H_0\mathbf {Q}_k \bar{\mathbf {v}}_0 \rbrace  }{ \frac{ 1}{ \sqrt{t_0} } } - \frac{ \mathbf {w}^H_0\mathbf {Q}_k \mathbf {w}_0 }{ \frac{ 1}{ {t_0} } } \frac{ 1}{ \sqrt{\tau _0} }- \frac{ \mathbf {w}^H_0\mathbf {Q}_k \mathbf {w}_0 }{ \frac{ 1}{ {\sqrt{ t_0}} } } \right) \nonumber \\&= 4 t_0| {\bar{\mathbf {q}}}^H_k \mathbf {w}_0 | ^2 \mathrm {Re}\lbrace  \mathbf {w}_0^H\mathbf {Q}_k {\bar{\mathbf {v}}}_0 \rbrace  - \frac{ 2( \mathbf {w}_0^H\mathbf {Q}_k \mathbf {w}_0 )^2t_0^{\frac{3}{2}} }{ \sqrt{\tau _0}} - (\mathbf {w}_0^H\mathbf {Q}_k\mathbf {w}_0)^2t_0 \nonumber \\& \triangleq f( {\bar{\mathbf {v}}}_0, \tau _0).\) 
